What are the solutions to the equation x^2-9x+7=0 ?

The solutions to the equation x^2-9x+7=0 are x=(9+sqrt(53))/2 ,x=(9-sqrt(53))/2

Find the zeros of x^2-9x+7=0

The zeros of x^2-9x+7=0 are x=(9+sqrt(53))/2 ,x=(9-sqrt(53))/2

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x^{2} - 9 x + 7 = 0

x = \frac{9 + 53}{2}, x = \frac{9 - 53}{2}

Decimal Notation

x = 8.14005 \dots, x = 0.85994 \dots

Solution steps

x^{2} - 9 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 7 = 53

x_{1, 2} = \frac{- ( - 9 ) \pm 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 53}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 53}{2 \cdot 1}

x = \frac{- ( - 9 ) + 53}{2 \cdot 1} : \frac{9 + 53}{2}

x = \frac{- ( - 9 ) - 53}{2 \cdot 1} : \frac{9 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 53}{2}, x = \frac{9 - 53}{2}

(x + 2) (2 x - 1) (3 x + 7) \geq 0

\frac{( x + 2 ) ( 2 x - 3 )}{( x - 3 ) ( x + 1 )} \leq 6

(p - 3) (p - 1) = 0

3 t^{2} - 3 t + 1 = 0

- 9.79 + 7.8 v = - 17.39 - 11.1 v + 18.4 v

What are the solutions to the equation x^2-9x+7=0 ?
The solutions to the equation x^2-9x+7=0 are x=(9+sqrt(53))/2 ,x=(9-sqrt(53))/2
Find the zeros of x^2-9x+7=0
The zeros of x^2-9x+7=0 are x=(9+sqrt(53))/2 ,x=(9-sqrt(53))/2